Math Study Notebook — Algebra 1 & Geometry

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Algebra 1 — 10 Core Problems

Q1 Linear Equations ⚠️ Tricky Sign! ★☆☆ Easy

📌 ISOLATE the variable — move numbers to one side, variable to the other. Do the SAME to BOTH sides!

✏️ Example (Similar)

Solve: \(2x + 3 = 9\)
Step 1: Subtract 3 → \(2x = 6\)
Step 2: Divide by 2 → \(x = 3\) ✓

Solve for \(x\):
\(3x - 7 = -1\)
Hint: Don't forget — subtracting a negative = adding!

📖 Step-by-Step Solution

\(3x - 7 = -1\)
Add 7 to both sides: \(3x = -1 + 7 = 6\)
Divide both sides by 3: \(x = 2\) ✓
Common Mistake: Many students forget that \(-1 + 7 = +6\), not \(-8\). Always watch the signs!

Q2 Slope ⚠️ Rise over Run! ★☆☆ Easy

📌 SLOPE = \(\frac{rise}{run} = \frac{y_2 - y_1}{x_2 - x_1}\) — always y on top, x on bottom!

✏️ Example (Similar)

Points \((1, 2)\) and \((3, 6)\)
Slope \(= \frac{6-2}{3-1} = \frac{4}{2} = 2\) ✓

What is the slope of a line passing through \((2, 1)\) and \((6, 9)\)?
Watch the order — keep the same point first in both numerator and denominator!

📖 Step-by-Step Solution

\(m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{9-1}{6-2} = \frac{8}{4} = 2\) ✓
Common Mistake: Flipping it as \(\frac{x_2-x_1}{y_2-y_1}\) gives \(\frac{1}{2}\) — always y-difference over x-difference!

Q3 Exponents ⚠️ Zero Power! ★☆☆ Easy

📌 ANYTHING to the power of ZERO = 1! (except 0⁰, which is undefined). \(a^0 = 1\)

✏️ Example (Similar)

\(5^0 = 1\), \(\quad (xyz)^0 = 1\), \(\quad (-7)^0 = 1\)

Simplify: \(4x^0 + 3\) (where \(x \neq 0\))
Be careful — is the exponent applied to 4 or just x?

📖 Step-by-Step Solution

\(x^0 = 1\), so \(4x^0 = 4 \times 1 = 4\)
Then \(4 + 3 = 7\) ✓
Common Mistake: Writing \((4x)^0 = 1\). The exponent is only on \(x\), not on 4!

Q4 Distributive Property ⚠️ Negative Distributor! ★★☆ Medium

📌 DISTRIBUTE to ALL terms inside. \(a(b+c) = ab + ac\). When negative: FLIP both signs inside!

✏️ Example (Similar)

\(-2(x - 3) = -2 \cdot x + (-2)(-3) = -2x + 6\)

Expand and simplify: \(-3(2x - 4) + x\)
Don't forget to distribute the negative sign to EVERY term!

📖 Step-by-Step Solution

\(-3(2x - 4) + x\)
Distribute: \(-6x + 12 + x\)
Combine like terms: \((-6+1)x + 12 = -5x + 12\) ✓
Common Mistake: Writing \(-3(-4) = -12\) (forgetting negative × negative = positive!)

Q5 Factoring ⚠️ Sum & Product! ★★☆ Medium

📌 Factor \(x^2 + bx + c\): Find two numbers that MULTIPLY to c and ADD to b. ✕ then +

✏️ Example (Similar)

Factor \(x^2 + 5x + 6\)
Need: \(_ \times _ = 6\) AND \(_ + _ = 5\) → 2 and 3
Answer: \((x+2)(x+3)\) ✓

Factor completely: \(x^2 - 5x + 6\)
Both numbers must be NEGATIVE (product positive, sum negative)!

📖 Step-by-Step Solution

Need: \(_ \times _ = +6\) AND \(_ + _ = -5\)
Try: \(-2 \times -3 = 6\) ✓ and \(-2 + (-3) = -5\) ✓
So: \((x-2)(x-3)\) ✓
Common Mistake: Using \((x+2)(x+3)\) — gives \(+5x\), not \(-5x\)!

Q6 Systems of Equations ⚠️ Substitution Trap! ★★☆ Medium

📌 SUBSTITUTION: isolate one variable → plug into the other equation → solve → back-substitute!

Solve the system:
\(\begin{cases} y = 2x + 1 \\ 3x + y = 16 \end{cases}\)
The first equation is already solved for y — use substitution!

📖 Step-by-Step Solution

Substitute \(y = 2x+1\) into \(3x + y = 16\):
\(3x + (2x+1) = 16\)
\(5x + 1 = 16 \Rightarrow 5x = 15 \Rightarrow x = 3\)
Back-substitute: \(y = 2(3)+1 = 7\) ✓

Q7 Inequalities ⚠️ FLIP when Dividing by Negative! ★★☆ Medium

📌 Multiply/divide by a NEGATIVE number → FLIP the inequality sign! \(<\) becomes \(>\)

✏️ Example (Similar)

Solve: \(-2x < 8\)
Divide by \(-2\) → FLIP: \(x > -4\) ✓

Solve: \(-4x + 2 \geq 18\)
You WILL divide by a negative — remember to flip!

📖 Step-by-Step Solution

\(-4x + 2 \geq 18\)
Subtract 2: \(-4x \geq 16\)
Divide by \(-4\) → FLIP \(\geq\) to \(\leq\): \(x \leq -4\) ✓
Common Mistake: Forgetting to flip the sign when dividing by \(-4\)!

Q8 Quadratic Formula ⚠️ Discriminant Check! ★★☆ Medium

📌 DISCRIMINANT \(= b^2 - 4ac\): Positive → 2 real roots, Zero → 1 root, Negative → No real roots

Using the quadratic formula, solve: \(x^2 - 6x + 9 = 0\)
Formula: \(\displaystyle x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
Check the discriminant first — what does it tell you?

📖 Step-by-Step Solution

\(a=1, b=-6, c=9\)
Discriminant: \((-6)^2 - 4(1)(9) = 36 - 36 = 0\)
Since discriminant \(= 0\), exactly ONE solution!
\(x = \frac{-(-6)}{2(1)} = \frac{6}{2} = 3\) ✓
Note: \(x^2-6x+9 = (x-3)^2\) — a perfect square!

Q9 Functions ⚠️ f(x) notation confusion! ★☆☆ Easy

📌 f(x) just means "plug in x". f(3) = replace every x with 3!

Given \(f(x) = 2x^2 - x + 4\), find \(f(-2)\).
(-2)² is POSITIVE! Be careful squaring negatives.

📖 Step-by-Step Solution

\(f(-2) = 2(-2)^2 - (-2) + 4\)
\(= 2(4) + 2 + 4\)
\(= 8 + 2 + 4 = 14\) ✓
Common Mistake: Writing \((-2)^2 = -4\). Squaring a negative ALWAYS gives a positive!

Q10 Slope-Intercept Form ⚠️ Which is m, which is b? ★★☆ Medium

📌 \(y = \)m\(x + \)b — m = slope (steepness), b = y-intercept (where it crosses y-axis)

A line has slope \(m = -\frac{2}{3}\) and passes through \((0, 5)\).
Which equation represents this line? What is \(f(3)\)?
The point (0, 5) gives you the y-intercept directly!

📖 Step-by-Step Solution

\(y = mx + b\) with \(m = -\frac{2}{3}\) and \(b = 5\):
\(y = -\frac{2}{3}x + 5\)
\(f(3) = -\frac{2}{3}(3) + 5 = -2 + 5 = 3\) ✓

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Geometry — 10 Core Problems

G1 Pythagorean Theorem ⚠️ Which side is c? ★☆☆ Easy

📌 a² + b² = c² — c is ALWAYS the HYPOTENUSE (the longest side, opposite the right angle)!

✏️ Example (Similar)

Legs: 3, 4. Find hypotenuse.
\(3^2 + 4^2 = c^2 \Rightarrow 9 + 16 = 25 \Rightarrow c = 5\) ✓

A right triangle has legs of length 5 and 12. What is the length of the hypotenuse?
This is a famous Pythagorean triple — 5, 12, __

📖 Step-by-Step Solution

\(a^2 + b^2 = c^2\)
\(5^2 + 12^2 = c^2\)
\(25 + 144 = 169\)
\(c = \sqrt{169} = 13\) ✓ (The 5-12-13 triple is very common on tests!)

G2 Triangle Angles ⚠️ Sum is ALWAYS 180°! ★☆☆ Easy

📌 TRIANGLE SUM: All 3 interior angles always add to 180°, no exceptions!

Two angles of a triangle are 47° and 63°. What is the third angle?
Subtract both known angles from 180°

📖 Step-by-Step Solution

Third angle \(= 180° - 47° - 63° = 180° - 110° = 70°\) ✓
Check: \(47 + 63 + 70 = 180\) ✓

G3 Circle — Area & Circumference ⚠️ r vs d Confusion! ★★☆ Medium

📌 AREA = \(\pi r^2\) (r squared), CIRCUMFERENCE = \(2\pi r\) (r not squared). Don't mix them up!

A circle has diameter 10. What is its area?
Diameter given → radius = diameter ÷ 2 first!

📖 Step-by-Step Solution

Diameter \(= 10\), so radius \(r = 5\)
Area \(= \pi r^2 = \pi(5)^2 = 25\pi\) ✓
Common Mistake: Using diameter in formula: \(\pi(10)^2 = 100\pi\) — wrong! Always halve to get radius first!

G4 Parallel Lines & Transversals ⚠️ Alternate vs Co-interior! ★★☆ Medium

📌 ALTERNATE interior = EQUAL (Z-shape). CO-INTERIOR = add to 180° (C-shape). CORRESPONDING = EQUAL (F-shape).

Two parallel lines are cut by a transversal. One angle is 115°. What is the measure of its co-interior (same-side interior) angle?
Co-interior angles are supplementary — they add up to 180°!

📖 Step-by-Step Solution

Co-interior angles add to 180°:
\(180° - 115° = 65°\) ✓
Common Mistake: Thinking co-interior = equal (that's alternate interior angles!). Co-interior = supplementary (= 180°).

G5 Volume of Cylinder ⚠️ Use r, not d! ★★☆ Medium

📌 V = πr²h — Base area (circle) × Height. Think: stack of circles!

A cylinder has radius 3 and height 7. What is its volume?
Leave answer in terms of π

📖 Step-by-Step Solution

\(V = \pi r^2 h = \pi (3)^2 (7) = \pi \cdot 9 \cdot 7 = 63\pi\) ✓
Common Mistake: Forgetting to square the radius: \(\pi \cdot 3 \cdot 7 = 21\pi\) — wrong!

G6 Special Right Triangles ⚠️ 30-60-90 Ratios! ★★☆ Medium

📌 30-60-90: sides = \(x\) : \(x\sqrt{3}\) : \(2x\). Short leg × 2 = hypotenuse. 45-45-90: \(x\) : \(x\) : \(x\sqrt{2}\).

In a 30-60-90 triangle, the short leg = 5. What is the hypotenuse?
In 30-60-90: hypotenuse = 2 × short leg. No calculator needed!

📖 Step-by-Step Solution

30-60-90 ratio: \(x : x\sqrt{3} : 2x\)
Short leg (opposite 30°) \(= x = 5\)
Hypotenuse \(= 2x = 2(5) = 10\) ✓
Long leg (opposite 60°) would be \(= 5\sqrt{3}\)

G7 Similarity & Proportions ⚠️ Match corresponding sides! ★★☆ Medium

📌 SIMILAR shapes: same shape, different size. Set up CORRESPONDING RATIOS and cross-multiply!

Triangle ABC ~ Triangle DEF. If \(AB = 4\), \(BC = 6\), and \(DE = 10\), find \(EF\).
Set up a proportion: \(\frac{AB}{DE} = \frac{BC}{EF}\)

📖 Step-by-Step Solution

\(\frac{AB}{DE} = \frac{BC}{EF}\)
\(\frac{4}{10} = \frac{6}{EF}\)
Cross-multiply: \(4 \cdot EF = 60\)
\(EF = 15\) ✓

G8 Exterior Angle Theorem ⚠️ Not 180° — it's a sum! ★★☆ Medium

📌 EXTERIOR ANGLE = sum of the TWO NON-ADJACENT interior angles. Think: "far away angles add up!"

An exterior angle of a triangle measures 110°. One of the non-adjacent interior angles is 65°. What is the other non-adjacent interior angle?
Exterior = remote interior 1 + remote interior 2

📖 Step-by-Step Solution

Exterior angle = sum of two non-adjacent interior angles:
\(110° = 65° + x\)
\(x = 110° - 65° = 45°\) ✓
Common Mistake: Using \(180° - 110° = 70°\). That gives the adjacent interior angle, not the remote ones!

G9 Midpoint Formula ⚠️ Average both coordinates! ★☆☆ Easy

📌 MIDPOINT = \(\left(\frac{x_1+x_2}{2},\ \frac{y_1+y_2}{2}\right)\) — just AVERAGE the x's and AVERAGE the y's!

Find the midpoint of the segment with endpoints \((-3, 4)\) and \((7, -2)\).
Add the x's and divide by 2. Same for y's.

📖 Step-by-Step Solution

\(M = \left(\frac{-3+7}{2},\ \frac{4+(-2)}{2}\right) = \left(\frac{4}{2},\ \frac{2}{2}\right) = (2, 1)\) ✓

G10 Surface Area of a Cone ⚠️ Slant height ≠ regular height! ★★★ Hard

📌 CONE SA = \(\pi r^2 + \pi r l\) — base circle PLUS lateral face. l = slant height (not the vertical height h!)

A cone has radius \(r = 6\) and slant height \(l = 10\). What is its total surface area?
Total SA = base (circle) + lateral face (sector). Leave answer in terms of π.

📖 Step-by-Step Solution

Total Surface Area \(= \pi r^2 + \pi r l\)
\(= \pi(6)^2 + \pi(6)(10)\)
\(= 36\pi + 60\pi = 96\pi\) ✓
Common Mistake: Using height instead of slant height! If given height, find slant height first using Pythagorean theorem: \(l = \sqrt{r^2 + h^2}\)

✏️ Keep practicing — you've got this! ✏️